
Unlike the setting of [1]}, the noise in derivative estimate \(\varepsilon _{k}\)  is biased, i.e., \(E[\varepsilon _{k}] \ne 0\) . Now, unrolling the recursion above and taking expectations, we obtain
\(E\Vert z_{n}\Vert ^2 &\le 3 E[\Vert z_0\Vert ^2]\prod \limits _{k=1}^{n}(1 - a_{k}M_{k-1})^2+ 3 E[\sum \limits _{k=1}^{n}[a_{k} \varepsilon _{k-1} \prod \limits _{j=k+1}^{n}(1 - a_{j}M_{j-1})]^2\\&\le 3E[\Vert z_0\Vert ^2]n^{-2\mu _2 c} + 3 \underbrace{\sum \limits _{k=1}^{n}\frac{c^2}{k^2} E[\varepsilon _{k-1}^2] (\mathcal {P}_{k+1:n})^2}_\text{I} + 3\underbrace{\sum \limits _{k\ne l}^{n}a_{k}a_{l}E[|\varepsilon _{l-1}|] E[|\varepsilon _{k-1}|] \mathcal {P}_{k+1:n}\mathcal {P}_{l+1:n}}_\text{II},\) 
